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Q. 104

Expert-verified
Found in: Page 487

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

# Show that the difference quotient for $f\left(x\right)=\mathrm{cos}x$is given byrole="math" localid="1646431168099" $\frac{f\left(x+h\right)-f\left(x\right)}{h}=\frac{\mathrm{cos}\left(x+h\right)-\mathrm{cos}\left(x\right)}{h}\phantom{\rule{0ex}{0ex}}=-\mathrm{sin}x·\frac{\mathrm{sin}h}{h}-\mathrm{cos}x·\frac{1-\mathrm{cos}h}{h}$

The difference quotient for $f\left(x\right)=\mathrm{cos}\left(x\right)$is determined by using the Sum formula for the cosine function as

$\frac{f\left(x+h\right)-f\left(x\right)}{h}=\frac{\mathrm{cos}\left(x+h\right)-\mathrm{cos}\left(x\right)}{h}\phantom{\rule{0ex}{0ex}}=\frac{\left(\mathrm{cos}x\right)\left(\mathrm{cos}h\right)-\left(\mathrm{sin}x\right)\left(\mathrm{sin}h\right)-\mathrm{cos}\left(x\right)}{h}\phantom{\rule{0ex}{0ex}}=-\mathrm{sin}x·\frac{\mathrm{sin}h}{h}-\mathrm{cos}x·\frac{1-\mathrm{cos}h}{h}$

See the step by step solution

## Step 1. Given data

The given function is

$f\left(x\right)=\mathrm{cos}x$

the equation that needs to prove is

$\frac{f\left(x+h\right)-f\left(x\right)}{h}=\frac{\mathrm{cos}\left(x+h\right)-\mathrm{cos}\left(x\right)}{h}\phantom{\rule{0ex}{0ex}}=-\mathrm{sin}x·\frac{\mathrm{sin}h}{h}-\mathrm{cos}x·\frac{1-\mathrm{cos}h}{h}$

## Step 2. Derivation

Use Sum formula for the cosine function

$\frac{f\left(x+h\right)-f\left(x\right)}{h}=\frac{\mathrm{cos}\left(x+h\right)-\mathrm{cos}\left(x\right)}{h}\phantom{\rule{0ex}{0ex}}=\frac{\left(\mathrm{cos}x\right)\left(\mathrm{cos}h\right)-\left(\mathrm{sin}x\right)\left(\mathrm{sin}h\right)-\mathrm{cos}\left(x\right)}{h}\phantom{\rule{0ex}{0ex}}=\frac{-\left(\mathrm{sin}x\right)\left(\mathrm{sin}h\right)}{h}-\frac{\mathrm{cos}\left(x\right)-\left(\mathrm{cos}x\right)\left(\mathrm{cos}h\right)}{h}\phantom{\rule{0ex}{0ex}}=-\left(\mathrm{sin}x\right)\frac{\left(\mathrm{sin}h\right)}{h}-\frac{\left(\mathrm{cos}x\right)\left(1-\left(\mathrm{cos}h\right)\right)}{h}\phantom{\rule{0ex}{0ex}}=-\mathrm{sin}x·\frac{\mathrm{sin}h}{h}-\mathrm{cos}x·\frac{1-\mathrm{cos}h}{h}$